Question

Danny (120kg) and June (50kg) are both at the opposite ends of a rope that they are connected by. They are standing on a frozen lake (frictionless). They are 15m apart when Danny starts pulling on the rope. This gives June an acceleration of 1.2m/s.

- What is Danny’s acceleration? Approximate where they will meeting relative to where Danny started from?

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Answer 1

The tension uniformly propagation through rope, hence same force acting on the both ends(both persons)

mass Danny M= 120 kg

mass of June m= 50 kg

acceleration of June a_J = 1.2 m/s^2

force acting on Danny =  force acting on June

Ma_D = ma_J

therefore acceleration of Danny a_{D =} ma_J / M = 50*1.2/120 = 0.5 m/s^2

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Center of mass remains stationary in absence of any external forces.
So the length of the rope can be divided in inverse ratio of the masses.

M / m=L₂ / L₁
L₂ / L₁ = 120/50 = 2.4

L₂ = 2.4 L₁

L₁ +  L₂ = 15 meters

L₁ + 2.4 L₁ = 15

therfore L₁ = 15/3.4 =4.412 meters

Therefore , they will meet from the initial position of Danny = 15- 4.412 = 10.588 meters